Transfer function algebraic manipulation

Click For Summary
SUMMARY

The discussion centers on the algebraic manipulation of the transfer function \( g_{p} = \frac{-1.43}{(s-1.399)(s+5.086)} \). Participants highlight the importance of correctly factoring the denominator and combining constants to achieve the desired form. A key insight is that the negative time constant, while indicative of instability, is permissible in the context of control systems. The final form of the transfer function requires careful attention to the constants involved, particularly noting that \( k_{p} \) does not equal -1.43.

PREREQUISITES
  • Understanding of transfer functions in control systems
  • Familiarity with algebraic manipulation techniques
  • Knowledge of stability concepts in control theory
  • Basic proficiency in Laplace transforms
NEXT STEPS
  • Study the implications of negative time constants in control systems
  • Learn about the stability criteria for transfer functions
  • Explore the process of factoring polynomials in the context of control theory
  • Investigate the significance of combining constants in transfer function representation
USEFUL FOR

Control engineers, students studying control systems, and anyone involved in the analysis and design of dynamic systems will benefit from this discussion.

gfd43tg
Gold Member
Messages
948
Reaction score
48

Homework Statement


upload_2015-11-7_14-2-6.png


Homework Equations

The Attempt at a Solution


I don't know if I am just stupid, but I can't see how to make this manipulation. I tried factoring the denominator

$$ g_{p} = \frac {-1.43}{(s-1.399)(s+5.086)} $$
Then take out -1.399 and 5.086
$$ g_{p} = \frac {-1.43}{-1.399(\frac {-1}{1.399}s + 1)5.086(\frac {1}{5.086}s+1)} $$
But I know this won't work, and I am not clever enough right now to come up with the right way to manipulate this thing.
 
Physics news on Phys.org
Maylis said:

Homework Statement


View attachment 91494

Homework Equations

The Attempt at a Solution


I don't know if I am just stupid, but I can't see how to make this manipulation. I tried factoring the denominator

$$ g_{p} = \frac {-1.43}{(s-1.399)(s+5.086)} $$
Then take out -1.399 and 5.086
$$ g_{p} = \frac {-1.43}{-1.399(\frac {-1}{1.399}s + 1)5.086(\frac {1}{5.086}s+1)} $$
But I know this won't work, and I am not clever enough right now to come up with the right way to manipulate this thing.

Your manipulations are correct; it is just that one of your constants ##\tau_i## is negative. If that is not allowed, there must be something wrong with the original ##g_p(s)##.
 
  • Like
Likes gfd43tg
The negative time constant is allowed, which is what makes it unstable. My algebra is weak, so for some reason I had it in my head that if I multiplied my result out, I would come up with something different from the original transfer function!
 
Maylis said:

Homework Statement


View attachment 91494

Homework Equations

The Attempt at a Solution


I don't know if I am just stupid, but I can't see how to make this manipulation. I tried factoring the denominator

$$ g_{p} = \frac {-1.43}{(s-1.399)(s+5.086)} $$
Then take out -1.399 and 5.086
$$ g_{p} = \frac {-1.43}{-1.399(\frac {-1}{1.399}s + 1)5.086(\frac {1}{5.086}s+1)} $$
But I know this won't work, and I am not clever enough right now to come up with the right way to manipulate this thing.
You've almost got the form of gp(s) that is specified, you just need to combine all the constants in the numerator and the denominator.

Hint: kp ≠ -1.43 in the final form.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
Replies
11
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 21 ·
Replies
21
Views
1K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
26
Views
4K